Thank you Murali, that does it for me.

2009/7/13 <Murali.MENON@fortisinvestments.com>

> This appears to work:
>
> > aapl <- as.xts(read.zoo(textConnection("2007-04-20,      90.97
> + 2007-04-23,      93.51
> + 2007-04-24,      93.24
> + 2007-04-25,      95.35
> + 2007-04-26,      98.84
> + 2007-04-27,      99.92
> + 2007-04-30,      99.80
> + 2007-05-01,      99.47
> + 2007-05-02,     100.39
> + 2007-05-03,     100.40
> + 2007-05-04,     100.81"), sep=","))
>
> > aapl[grep("2007-04-30", index(aapl)) - 1]
>            [,1]
> 2007-04-27 99.92
>
> Murali
>
> -----Original Message-----
> From: r-sig-finance-bounces@stat.math.ethz.ch
> [mailto:r-sig-finance-bounces@stat.math.ethz.ch] On Behalf Of Mark
> Breman
> Sent: 13 July 2009 07:57
> To: r-sig-finance@stat.math.ethz.ch
> Subject: [R-SIG-Finance] xts question: how to get previous row?
>
> I have a really basic question but I can't find an answer for it.
>
> Supose I have this xts timeseries called aapl:
>           AAPL.Close
> 2007-04-20      90.97
> 2007-04-23      93.51
> 2007-04-24      93.24
> 2007-04-25      95.35
> 2007-04-26      98.84
> 2007-04-27      99.92
> 2007-04-30      99.80
> 2007-05-01      99.47
> 2007-05-02     100.39
> 2007-05-03     100.40
> 2007-05-04     100.81
>
> and supose I have a reference to a row in this series (i.e.
> aapl["2007-04-30"]), what is the easiest way to select the previous row
> (i.e. the row with index 2007-04-27) from the series?
>
> Thanks,
>
> -Mark-
>
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>
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